Hi heathtech,
As expected, your optic switch power supply is rated for 5V operation only. Now you have connected it to 15V by mistake, you might have damaged it already.
Please get the datasheet of the sensor from the following site:
**broken link removed**
After careful thoughts, the use of toggle flip-flop for this particular operation is not foolproof and an extra pulse from the sensor would result in the circuit failed to operate at all.
What I propose now is using the 4013 in the set-reset mode. This has more advantage over the toggle mode as it can now survive several pulses from the sensor and still work correctly.
Circuit operates as follow:
Sensor output is normally LOW when light is blocked. Both AND gate 4081 outputs LOW. Let assume at this moment 4013 Q=LOW and /Q=HIGH. The hole of the rotating disc is opposite the sensor. C1 has already charged up to a voltage close to +5V via R3. Capacitor C2 voltage is LOW because Q is LOW.
When the hole of the disc moves across the sensor, the output of the sensor goes HIGH. U1A now ouputs HIGH and SET the flipflop so that 4013 Q=HIGH and /Q=LOW.
This switches off the AND gate U1A and extra pulses, if any, coming from the sensor has no effect on the 4013.
The motor now reverse direction and drive the hole of the disc away from the sensor. The sensor output returns to LOW.
Meanwhile, C2 now is charging up via R4. The timing of R4/C2 and R3/C1 is such that the voltage rise at the capacitor would only be high enough sometimes after the sensor output has returned to LOW. Without know the rotating speed of the disc, the value of C I put down is a wild guess but it is not critical to the operation of the circuit as long as it is completely charged up before the "hole" has moved towards the slot of the sensor.
The next time the sensor goes HIGH, a RESET operation would be carried out and the cycle repeats.
Heathtech you should be able to figure out the full schematic from the above. Remember to follow the good practice of grounding all unused CMOS gate inputs and use additional ceramic capacitors for the power supply bypass.
Good luck to you and please let us know whether the problem is solved.